Question

i=sqrt(-1)
A=(i^3)(i^29)(i^208)

Answer 1

\[i^3\cdot i^{29}\cdot i^{208}=i^{3+29+208}=i^{240}\]Do you know anything about powers of i, and how they cycle?

Answer 2

goes like
\[i^0=1\]
\[i^1=i\]
\[i^2=-1\]
\[i^3=-1\]
\[i^4=1\] so take the integer remainder when you divide the exponent by 4, and you will get one of the above 4 possibilities

Answer 3

since 4 divides 240 evenly you get 1

Answer 4

okay thanks

Answer 5

yw